Transforming Triangles with ABC Matrix

[Lonely Lemon]

Homework Statement

Write down 3x3 matrices A, B, C such that when the vectors in R2 are expressed in homogeneous coordinates, the product ABC first translates vectors by (-1, 2), then reflects them about the line y=-x and finally scales them by 2. using your matrix ABC, determine the image of the triangle with vertices (0,0), (1, 2), (2, 3).

Homework Equations

The Attempt at a Solution

I've already figured out A, B and C and the ABC matrix as [0 -2 -4; -2 0 2; 0 0 1] (each division separated by ';' being a column of the ABC matrix). But I can't figure out how I'm supposed to apply this to the vertices?

The answers are (-4, 2), (-8, 0), (-10, -2)

What should I multiply my ABC matrix by to see what happens to the vertices?

 

[CompuChip]

What should I multiply my ABC matrix by to see what happens to the vertices?

You should multiply it by the vertices, written as vectors.
Of course, since you have written ABC down in a homogeneous coordinate system, you will also have to write your vertices in homogeneous coordinates first (such that they are represented by 3-vectors rather than 2-vectors).

 

[Lonely Lemon]

I did that, my homogeneous vector matrix for the triangle is, call it P:

P = [0 0 1; 1 2 1; 2 3 1]

But if I multiply the ABC matrix by P such that I get [0 -2 -4; 2 0 2; 0 0 1][0 0 1; 1 2 1; 2 3 1] I end up with [0 0 1; -4 -2 1; -6 -4 -1] which is patently wrong...

 

[CompuChip]

Try multiplying the vertices one by one, i.e. calculate [0 -2 -4; 2 0 2; 0 0 1][0 0 1], [0 -2 -4; 2 0 2; 0 0 1][1 2 1] and [0 -2 -4; 2 0 2; 0 0 1][2 3 1] where the vectors are column vectors, of course.

 

[Lonely Lemon]

That's still not giving me the correct result... If I go [0 -2 -4; -2 0 2; 0 0 1][0 0 1] for example [0 0 1] just shoots back out again, which can't be right when there is a translation involved

 

[CompuChip]

Hmm, now I got confused by the textual representation... is the [0 -2 -4] the first row, or the first column. What happens if you reverse that (i.e. transpose your matrix).
I did get the correct results using your matrix yesterday, namely, but I might have transposed it somewhere along.

 

[Lonely Lemon]

Ah, you're right! My textual representation did read [0 -2 -4] as a column, but if I do transpose the matrix I end up with the correct linear transformation.

I can see how the answer comes about now, but I still don't see why I need to transpose my transformation matrix? Can you explain?

 

[CompuChip]

I can't tell you exactly without seeing the whole calculation, but generally I can tell you this: if you have a basis for your space, and a transformation A on it, then the columns of A are the transformed basis vectors of your space.
Actually they have to be, if you consider a transformation from n to m dimensions: if A multiplies an n-dimensional vector to produce an m-dimensional one, then A has to have n columns with m rows, not vice versa. And if you take [1 0 0 ... 0], [0 1 0 0 ... 0], [0 0 1 0 ... 0], ..., [0 0 ... 0 1] as a basis and multiply with A, you should get the transformed basis (which is again a basis if A is an invertible linear transformation), and you will see that you get precisely the columns of A in this case.

In this case, if you pick your favorite vector and work out by hand what it transforms to, you can easily see what form of A should be the correct one.